Prove $\frac{d \ln(y)}{d \ln(x)} = \frac{dy}{dx} \frac{x}{y}$ using limits I see this equation used again and again in economics but I really can't find a rigorous limits based proof. 
I have that
$$\frac{d\ln(y)}{d\ln(x)} =  \lim_{h \to 0}\frac{\ln(f(\ln(x)+h))-\ln f(\ln(x))}{h}$$
Not sure how to finish. Thanks
 A: Let $z = \log x$, i.e. $x = \exp(z)$, and $y = y(x)$ be a smooth and nonzero enough function of $x$; then
$$\frac{d \log y(z)}{dz} = \frac{1}{y(z)} y'(z) = \frac{1}{y(z)} y'(x(z))x'(z) = \frac{dy}{dx} \frac{x}{y}$$ where we use the chain rule twice.
A: Implicit assumptions in your question:  $y = f(x)$, $x > 0$, $y > 0$.  With these, the following calculation is valid:
$$
\begin{align}
\frac{d \ln y}{d \ln x} &= \frac{d \ln y}{dx} \cdot \left(\frac{d \ln x}{dx}\right)^{-1} \\
&= \frac{1}{y}\frac{dy}{dx} \cdot \left( \frac{1}{x} \right)^{-1} \\
&= \frac{x}{y} \frac{dy}{dx}.
\end{align}
$$
A: Here is the limit solution, I thought of how it would work. It is true for $g$ differentiable with a nonzero derivative that
$$\frac{df(x)}{dg(x)}= \frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}$$
Proof: the RHS =  $$ \lim_{t \to t_0} \frac{f(t)-f(t_0)}{g(t)-g(t_0)}=\lim_{t \to t_0} \frac{{f(g^{-1}(g(t)))-f(g^{-1}(g(t))})}{g(t)-g(t_0)}$$
which is the definition of the derivative of f at t with respect to g(t). To see this, note that if u=g(x), $$ \frac{df}{du}= \frac{{f(g^{-1}(u))-f(g^{-1}(u_0)})}{g(t)-g(t_0)} = \frac{{f(g^{-1}(g(t)))-f(g^{-1}(g(t))})}{g(t)-g(t_0)}$$
The result desired follows from this formula.
